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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof:</dfn> Suppose that <span class="process-math">\(y=\phi(x)\)</span> is any solution of (<a href="" class="xref" data-knowl="./knowl/eq3_6.html" title="Equation 3.2.2">(3.2.2)</a>), we need to show that <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> can be found such that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\phi(x)=C_1 y_1(x)+C_2 y_2(x).
\end{equation*}
</div>
<p class="continuation">Denote <span class="process-math">\(\phi(x_0)=a_0\text{,}\)</span> <span class="process-math">\(\phi^{\prime}(x_0)=a_1\text{.}\)</span> Then, the initial value problem</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation}
y^{\prime \prime}+p(x) y^{\prime}+q(x)y=0,\quad y(x_0)=a_0, \quad y^{\prime}(x_0)=a_1\tag{3.2.4}
\end{equation}
</div>
<p class="continuation">has a unique solution according to the theorem on the uniqueness and existence. Then <span class="process-math">\(y=\phi(x)\)</span> should be this unique solution (it satisfies both the ODE and initial conditions). On the other hand, we know that for the function</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\bar{\phi}(x)=C_1 y_1(x)+C_2 y_2(x)
\end{equation*}
</div>
<p class="continuation">to satisfy (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>), we only have to choose</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
C_1=\frac{\left| \begin{array}{cc} a_0 &amp; y_2(x_0)\\ a_1  &amp; y_2^{\prime}(x_0) \end{array}    \right|}{W_0},\quad C_2=\frac{\left| \begin{array}{cc} y_1(x_0) &amp; a_0 \\ y_1^{\prime}(x_0) &amp;a_1 \end{array}    \right|}{W_0},
\end{equation*}
</div>
<p class="continuation">where</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation}
W_0=\left| \begin{array}{cc} y_1(x_0) &amp; y_2(x_0) \\ y_1^{\prime}(x_0) &amp; y_2^{\prime}(x_0) \end{array}    \right|.\tag{3.2.5}
\end{equation}
</div>
<p class="continuation">Since <span class="process-math">\(y_1(x)\)</span> and <span class="process-math">\(y_2(x)\)</span> are a fundamental set of solutions as shown by (<a href="" class="xref" data-knowl="./knowl/eq3_5_5.html" title="Equation 3.2.3">(3.2.3)</a>), <span class="process-math">\(W_0\)</span> in (<a href="" class="xref" data-knowl="./knowl/eq3_8.html" title="Equation 3.2.5">(3.2.5)</a>) is nonzero. So <span class="process-math">\(C_1\)</span> and <span class="process-math">\(C_2\)</span> are uniquely determined. Therefore <span class="process-math">\(\bar{\phi}(x)\)</span> is also a solution to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>). We know both <span class="process-math">\(\bar{\phi}(x)\)</span> and <span class="process-math">\(\phi(x)\)</span> are solutions to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq3_7.html" title="Equation 3.2.4">(3.2.4)</a>) while this initial value problem has a unique solution. Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_6.html ./knowl/eq3_7.html ./knowl/eq3_5_5.html ./knowl/eq3_8.html ./knowl/eq3_7.html ./knowl/eq3_7.html">
\begin{equation*}
\phi(x)=\bar{\phi}(x)=C_1 y_1(x)+C_2 y_2(x).
\end{equation*}
</div>
<p class="continuation">So <span class="process-math">\(y(x)=C_1 y_1(x)+C_2 y_2(x)\)</span> is the general solution.</p>
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